Find the temperature at which 2 moles of `SO_2` will occupy a volume of 10 litre at a pressure of 15 atm. a=6.71` atm litre^2mol^(−2)` ;b=0.0564 `litre mol^(−1)` .

Options

A. 919
B. 349
C. 489
D. None of these

Explanation

To find the temperature at which 2 moles of \( SO_2 \) will occupy a volume of 10 liters at a pressure of 15 atm using the Van der Waals equation, we will follow these steps:

### Step 1: Write the Van der Waals equation
The Van der Waals equation is given by:
\[ \left( P + \frac{a n^2}{V^2} \right) (V - nb) = nRT \]
Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles
- \( R \) = ideal gas constant (0.0821 L atm K\(^{-1}\) mol\(^{-1}\))
- \( a \) and \( b \) = Van der Waals constants for the gas

### Step 2: Substitute the known values into the equation
Given:
- \( P = 15 \) atm
- \( n = 2 \) moles
- \( V = 10 \) liters
- \( a = 6.71 \) atm L\(^2\) mol\(^{-2}\)
- \( b = 0.0564 \) L mol\(^{-1}\)

Substituting these values into the Van der Waals equation:
\[ \left( 15 + \frac{6.71 \times 2^2}{10^2} \right) (10 - 2 \times 0.0564) = 2RT \]

### Step 3: Calculate the left side of the equation
First, calculate \( \frac{6.71 \times 2^2}{10^2} \):
\[ \frac{6.71 \times 4}{100} = \frac{26.84}{100} = 0.2684 \]
Now substitute this back into the equation:
\[ \left( 15 + 0.2684 \right) (10 - 0.1128) = 2RT \]
Calculate \( 15 + 0.2684 \):
\[ 15.2684 \]
Now calculate \( 10 - 0.1128 \):
\[ 9.8872 \]
So, the equation becomes:
\[ 15.2684 \times 9.8872 = 2RT \]

### Step 4: Calculate the product
Now calculate \( 15.2684 \times 9.8872 \):
\[ 15.2684 \times 9.8872 \approx 151.56 \]
So we have:
\[ 151.56 = 2RT \]

### Step 5: Substitute \( R \) and solve for \( T \)
We know \( R = 0.0821 \) L atm K\(^{-1}\) mol\(^{-1}\). Therefore:
\[ 151.56 = 2 \times 0.0821 \times T \]
This simplifies to:
\[ 151.56 = 0.1642T \]
Now, solve for \( T \):
\[ T = \frac{151.56}{0.1642} \approx 922.5 \text{ K} \]

### Step 6: Final answer
Thus, the temperature at which 2 moles of \( SO_2 \) will occupy a volume of 10 liters at a pressure of 15 atm is approximately:
\[ T \approx 922.5 \text{ K} \]